CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Apparent Depth

An object at ...

Question

An object at the bottom of a beaker filled with a liquid up to height of 10 cm. If the refractive index of liquid w.r.t. air ( ^aµ_w ) is 4/3, find the apparent depth of the object.

Open in App

Solution

Real depth = 10 cm
^aµ_w = 4/3
µ = Real depth / Apparent depth
4/3 = 10 / Apparent depth
Apparent depth = 10 X 3 / 4
= 30 / 4
= 7. 5 cm (Ans.)

Suggest Corrections

thumbs-up

9

Similar questions

Q. A vessel of depth

2 h

is half filled with a liquid of refractive index

2 \sqrt{2}

and the upper half with another liquid of refractive index

\sqrt{2}

. The liquids are immiscible. The apparent depth of an object which has a height

\frac{h}{2}

from the bottom of the vessel inside the liquid as seen from the top of the vessel is

Q. A real image of an object is formed by a convex lens at the bottom of an empty beaker. The beaker is now filled with a liquid of refractive index 1.4 to a depth of 7 cm. In order to get the image again at the bottom, the beaker should be moved

Q. A vessel of depth d filled with a liquid of refractive index

μ_{1}

up to half its depth and the remaining space is filled with a liquid of refractive index

μ_{2}

. The apparent depth while seeing normal to the
free surface of the liquid is

Q. A bucket of total height 60 cm is half filled with a liquid of refractive index 1.5 and half with another liquid of refractive Index 2. The apparent depth of the bucket for an observer directly above the bucket is

Q. A vessel of depth 2d cm is half filled with a liquid of refractive index

μ_{1}

and the upper half with a liquid of refractive index

μ_{2}

. The apparent depth of the vessel seen perpendicularly is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Apparent Depth

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard X Physics

Join BYJU'S Learning Program

CrossIcon